$\dfrac{ 9e + 9f }{ -9 } = \dfrac{ e + g }{ 5 }$ Solve for $e$.
Solution: Multiply both sides by the left denominator. $\dfrac{ 9e + 9f }{ -{9} } = \dfrac{ e + g }{ 5 }$ $-{9} \cdot \dfrac{ 9e + 9f }{ -{9} } = -{9} \cdot \dfrac{ e + g }{ 5 }$ $9e + 9f = -{9} \cdot \dfrac { e + g }{ 5 }$ Multiply both sides by the right denominator. $9e + 9f = -9 \cdot \dfrac{ e + g }{ {5} }$ ${5} \cdot \left( 9e + 9f \right) = {5} \cdot -9 \cdot \dfrac{ e + g }{ {5} }$ ${5} \cdot \left( 9e + 9f \right) = -9 \cdot \left( e + g \right)$ Distribute both sides ${5} \cdot \left( 9e + 9f \right) = -{9} \cdot \left( e + g \right)$ ${45}e + {45}f = -{9}e - {9}g$ Combine $e$ terms on the left. ${45e} + 45f = -{9e} - 9g$ ${54e} + 45f = -9g$ Move the $f$ term to the right. $54e + {45f} = -9g$ $54e = -9g - {45f}$ Isolate $e$ by dividing both sides by its coefficient. ${54}e = -9g - 45f$ $e = \dfrac{ -9g - 45f }{ {54} }$ All of these terms are divisible by $9$ $e = \dfrac{ -{1}g - {5}f }{ {6} }$